Problem: Let $a(x)=x^5+2x^4-x^3+2$, and $b(x)=x^3+3$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{x^5+2x^4-x^3+2}{x^3+3}$ : First, we divide ${x^3}$ into ${x^5}$ and get ${x^2}$ : $ \hphantom{1567|1.4} {x^2}\\ {{{x^3}+3}}|\overline{{x^5}+2x^4-x^3+0x^2+2}\\ \hphantom{37....|}\llap{-}\underline{(x^5+0x^4+0x^3+3x^2)}\\ \hphantom{37|3........}+2x^4-x^3-3x^2\\ $ [What did we do here?] Next, we divide ${x^3}$ into ${2x^4}$ to get ${+2x}$, and continue doing this until we find the quotient: $ \hphantom{1567|14} {x^2 \ {+ \ 2x}\ -1}\\ {{{x^3}+3}}|\overline{x^5+2x^4-x^3+0x^2+0x+2}\\ \hphantom{37....|}\llap{-}\underline{(x^5+0x^4+0x^3+3x^2)}\\ \hphantom{37|3........}{+2x^4}-x^3-3x^2+0x\\ \hphantom{37...........|}\llap{-}\underline{(2x^4+0x^3+0x^2+6x)}\\ \hphantom{37|3....................}{-x^3-3x^2-6x+2}\\ \hphantom{3788888888888.|}\llap{-}\underline{(-x^3+0x^2+0x-3)}\\ \hphantom{37|3...............9999.....}{-3x^2-6x+5}\\ $ [What did we do here?] The process stops here because $x^3+3$ is a polynomial of the third degree and $-3x^2-6x+5$ is a polynomial of the second degree. So it follows that ${r(x)}={-3x^2-6x+5}$, ${q(x)}={x^2+2x-1}$, and $ \dfrac{x^5+2x^4-x^3+2}{x^3+3}={x^2+2x-1}+\dfrac{{-3x^2-6x+5}}{x^3+3}$ To conclude, $q(x)=x^2+2x-1$ $r(x)=-3x^2-6x+5$